优优班--学霸训练营 > 知识点挑题
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            • 1.
              计算:\((\tan 60^{\circ})^{-1}× \sqrt { \dfrac {3}{4}}-|- \dfrac {1}{2}|+23×0.125\).
            • 2.

              计算:\(2\sqrt{8}\div \sqrt{\dfrac{1}{2}}\times \sqrt{18}\).

            • 3.
              先化简,再求值:\( \dfrac {x^{2}-y^{2}}{x}÷( \dfrac {2xy-y^{2}}{x}-x)\),其中,\(x= \sqrt {3}+2\),\(y= \sqrt {3}-2\).
            • 4.
              已知实数\(a\)、\(b\)在数轴上的对应点如图所示,化简\( \sqrt {a^{2}}+|a+b|+| \sqrt {2}-a|- \sqrt {(b- \sqrt {2})^{2}}\)
            • 5.

              计算:\(2\sin 45{}^\circ -\left| -5 \right|+{{(\dfrac{1}{3}+\sqrt{3})}^{0}}-\sqrt{18}\).

            • 6.

              计算:\( \sqrt{18}−{(π−2017)}^{0}+|1− \sqrt{2}|+ \sqrt[3]{27} \).

            • 7.
              若\(x\),\(y\)为实数,且\(y < \sqrt {3-x}+ \sqrt {x-3}+2\),试化简:\(x^{2}+|y-2|- \sqrt {y^{2}-6y+9}\).
            • 8.
              观察下列各式:\( \dfrac {1}{ \sqrt {2}+1}= \dfrac {1×( \sqrt {2}-1)}{( \sqrt {2}+1)( \sqrt {2}-1)}= \dfrac { \sqrt {2}-1}{2-1}= \sqrt {2}-1\),同理:\( \dfrac {1}{ \sqrt {4}+ \sqrt {3}}=…= \sqrt {4}- \sqrt {3}\),\(…\)从计算结果中找出规律,并利用这一规律计算:\(( \dfrac {1}{ \sqrt {2}+1}+ \dfrac {1}{ \sqrt {3}+ \sqrt {2}}+ \dfrac {1}{ \sqrt {4}+ \sqrt {3}}+…+ \dfrac {1}{ \sqrt {2018}+ \sqrt {2017}})( \sqrt {208}+1)\)
            • 9.

              计算:\(2\sin 30^{\circ}+{{(\dfrac{1}{3})}^{-1}}+{{(4-\pi )}^{0}}+\sqrt{8}.\)

            • 10.

              计算:\(\sqrt{18}+\left| 1-\sqrt{2} \right|-2\cos 45{}^\circ +{{\left( \dfrac{1}{3} \right)}^{-1}}\).

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